# Thread: Quick question

1. ## Quick question

If $\displaystyle \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right)$

What is $\displaystyle \left<x^2\right>$?

2. Originally Posted by Captcha
If $\displaystyle \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right)$

What is $\displaystyle \left<x^2\right>$?
so square both sides of your initial equation
$\displaystyle (x)^2 = \frac{1}{n^2}\left(x_{1}+x_{2}+...+x_{n}\right)^2$

and evaluate further if you know what the x1,x2,....etc are

3. Originally Posted by purebladeknight
so square both sides of your initial equation
$\displaystyle (x)^2 = \frac{1}{n^2}\left(x_{1}+x_{2}+...+x_{n}\right)^2$

and evaluate further if you know what the x1,x2,....etc are
It's not $\displaystyle \left<x\right>^2$ I want to find. It's $\displaystyle \left<x^2\right>$.

Basically I'm trying to solve for $\displaystyle x$ the equation:

$\displaystyle 2.61 = \sqrt{\left<x^2\right>-\left<x\right>^2}$ to find the value of $\displaystyle x$, where $\displaystyle 2.61$ is the $\displaystyle \text{standard deviation}$ of the data $\displaystyle 8,7, 4, 10, 6, x$.

I know that $\displaystyle \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right)$.

$\displaystyle \Rightarrow \left<x^2\right> = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2$.

So $\displaystyle 2.61 = \sqrt{\left<x^2\right>-\frac{1}{36}\left(35+x\right)^2}$

But I've no idea how to get $\displaystyle \left<x^2\right>$.

4. Originally Posted by Captcha
It's not $\displaystyle \left<x\right>^2$ I want to find. It's $\displaystyle \left<x^2\right>$.

Basically I'm trying to solve for $\displaystyle x$ the equation:

$\displaystyle 2.61 = \sqrt{\left<x^2\right>-\left<x\right>^2}$ to find the value of $\displaystyle x$, where $\displaystyle 2.61$ is the $\displaystyle \text{standard deviation}$ of the data $\displaystyle 8,7, 4, 10, 6, x$.

I know that $\displaystyle \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right)$.

$\displaystyle \Rightarrow \left<x^2\right> = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2$.

So $\displaystyle 2.61 = \sqrt{\left<x^2\right>-\frac{1}{36}\left(35+x\right)^2}$

But I've no idea how to get $\displaystyle \left<x^2\right>$.
$\displaystyle s = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \overline{X} }$.

5. Originally Posted by mr fantastic
$\displaystyle s = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \overline{X} }$.
$\displaystyle \left<x^2\right> = \frac{\displaystyle\sum_{i=1}^n x_i^2}{n}?$

6. Just tell me how to calculate, or give me definition of, $\displaystyle \left<x^2\right>$ and I will take care of the rest.

7. Originally Posted by Captcha
Just tell me how to calculate, or give me definition of, $\displaystyle \left<x^2\right>$ and I will take care of the rest.
Well, I gave you a formula that will do the job of efficiently answering the question you're working on. But, if you insist:

$\displaystyle <X^2> = \frac{x_1^2 + x_2^2 + x_3^2 + .... + x_n^2}{n}$.

8. Originally Posted by mr fantastic
Well, I gave you a formula that will do the job of efficiently answering the question you're working on. But, if you insist:

$\displaystyle <X^2> = \frac{x_1^2 + x_2^2 + x_3^2 + .... + x_n^2}{n}$.
Thanks a million. You are Super Fantastic!